Certainly! Let's solve the quadratic equation that models Jace's height above the water, given by:
[tex]\[
h = -16t^2 + t + 7
\][/tex]
We need to determine the time [tex]\( t \)[/tex] when Jace hits the water. This happens when his height [tex]\( h \)[/tex] is zero:
[tex]\[
0 = -16t^2 + t + 7
\][/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
In our quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = 7 \)[/tex]
Let's start by calculating the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[
\Delta = b^2 - 4ac = 1^2 - 4(-16)(7) = 1 + 448 = 449
\][/tex]
Next, we compute the solutions for [tex]\( t \)[/tex]:
[tex]\[
t_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-1 \pm \sqrt{449}}{-32}
\][/tex]
Let's determine the two possible values of [tex]\( t \)[/tex]:
1. For [tex]\( t_1 \)[/tex]:
[tex]\[
t_1 = \frac{-1 - \sqrt{449}}{-32} \approx \frac{-1 - 21.1896201004171}{-32} \approx 0.6934256281380341
\][/tex]
2. For [tex]\( t_2 \)[/tex]:
[tex]\[
t_2 = \frac{-1 + \sqrt{449}}{-32} \approx \frac{-1 + 21.1896201004171}{-32} \approx -0.6309256281380341
\][/tex]
Since time [tex]\( t \)[/tex] cannot be negative, we select the positive value:
[tex]\[
t = 0.6934256281380341 \text{ seconds}
\][/tex]
Therefore, it takes approximately [tex]\( 0.693 \)[/tex] seconds for Jace to hit the water.
[tex]\[
t = 0.693 \text{ seconds}
\][/tex]
